applied energy systems- hydrostatic fluid power transmission and

8/17/2019 Applied Energy Systems- Hydrostatic Fluid Power Transmission And

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Reynolds Number .......................................................................................... 49 Head Loss Due to Friction .......................................................................... 50 Flow through Valves and Fittings ........................................................... 54 Flow Around Bends and Elbows ............................................................. 54 Equivalent Pipe Length ............................................................................... 54

Pressure Loss and Pumping Power ....................................................... 55 Summary of Key Formulae for Pipe Flow ........................................... 56 Chapter 10. Valves and Heat Exchanger Flow Paths .................... 61 Valves .................................................................................................................. 61 Head Loss Coefficients for open Valves ............................................... 63 Heat Exchangers ............................................................................................. 64 Pressure Drops or Head Losses in Heat Exchangers ...................... 68

Important Formulae ......................................................................................... 70

Newton’s Viscosity Equation .................................................................... 70 Kinematic viscosity ....................................................................................... 70 Newton’s Second Law Applied to S.I. Units ........................................ 71 Pressure ............................................................................................................. 71 Pressure and Head ........................................................................................ 71 Compressibility and Bulk Modulus ........................................................ 72 Force Produced by a Hydraulic Cylinder ............................................. 72 Displacement of a Hydraulic Piston Pump ......................................... 72 Ideal Flow Rate of a Positive Displacement Pump .......................... 73 Efficiency of a Hydraulic Pump ................................................................ 73

Area of a Circle ................................................................................................ 74 Mass Flow Rate ............................................................................................... 74 Volume Flow Rate .......................................................................................... 74 Reynolds Number .......................................................................................... 74 The Darcy-Weisbach Equation ................................................................ 75 Friction Factor for Laminar Flow ........................................................... 76 Head Loss Coefficient for Fittings and Features............................... 76

Glossary .................................................................................................................. 77


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NomenclatureSymbol Description SI units

area m

vc area at vena contracta m o area of orifice m acceleration m s constant constant

distance or diameter meq equivalent diameter m force N friction factor acceleration due to gravity m s ℎ head, height or depth mℎ head loss due to friction m bulk modulus Pa head loss coefficient surface roughness mL length dimension

length mM mass dimension mass kg mass flow rate kg s cyl number of cylinders


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Symbol Description SI units Power W

pressure Nm

orPa volume flow rate ms Re Reynolds numberSI Système International,International System distance m

T time dimension time s volume m velocity m s velocity at position m s work J distance (usuallyhorizontal) m distance (usually vertical) mGreek letters compressibility Pa−1 (beta)Δ indicates an increment of

the following variable e.g.Δ is a difference betweentwo values of (uppercasedelta)Δhx pressure drop across a heatexchanger Pa


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Chapter 1.

IntroductionApplied Energy Systems modules have developed and evolved at

the Dublin Institute of Technology (DIT) within the framework ofthe Bachelor of Engineering Technology degree and theprogrammes that preceded it. The present work is a basictextbook or set of notes for such a module addressing basic FluidMechanics for incompressible fluids.Currently ‘energy’ is very topical. Sustainable energy andrenewable energy are buzz words. There is a widespreadappreciation that energy must be used efficiently and a generalwillingness to make use of all forms of energy that are natural andfree. Likewise there is a general desire to avoid, where possible,making use of forms of energy that are finite (i.e. already stored infixed quantities on the planet Earth) or that damage theenvironment when they are used. This present work fits into thatgeneral picture. The underlying principles and scientificknowledge that relate to energy use are common, whether energyis being used well or ineffectively and irrespective of the source ofthe energy. The use of fluids, e.g. gases like air or natural gas orliquids like water or gasoline, forms a major sub-category ofenergy systems generally.An interesting approach was used by those who developed anddelivered the Applied Energy Systems module for fluids at the DITin the past. Rather than presenting a course that attempted tocover Fluid Mechanics in a very broad and general way, theyidentified and selected two practical areas in the use of fluids:hydrostatic power transmission systems and the flow of fluidsthrough pipes and fittings. This selection allowed the two areas tobe presented in some detail, but it also allowed them to be sampleapplication areas that would enable the learners to put theprinciples of fluid mechanics into context, while developing theirknowledge and understanding in a way that would prepare them

for applying the same and similar principles to a much broaderrange of practical applications in the future.Hydrostatic power transmission systems are employed in a hugerange of heavy and light machinery used in the constructionindustry. In manufacturing industry, hydraulic power systems arecommon, e.g. for the operation of presses that can apply hugeforces to form work pieces or to allow the parts of moulding diesto be held firmly together and subsequently separated quickly.


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The principles of hydraulic power transmission are applied invehicles such as cars, trains, ships and aircraft. Furthermore,many of the principles of hydraulic power transmission also applyin situations where pressures act but there is not necessarilytransmission of power. For instance, a submarine is subjected tohydrostatic pressure.The flow of fluid in pipes is such a common phenomenon that wetake it completely for granted. Air flows into and out of our lungsthrough our bronchial tubes and our blood flows through pipesthat are usually called veins and arteries. Natural gas, oil andwater are just three fluids that commonly flow through pipes. It isa simple fact that energy is dissipated, or transformed to adifferent form, whenever a fluid flows. Mankind needs to useenergy to cause fluids to flow. While the present work is limited tothe application area of flow of fluids in pipes and fittings, the verysame principles are also common to other application areas suchas wind turbines, aerodynamics and biomedical devices such asartificial hearts.Learners who have ambition to work with the most exciting andinnovative ways of harnessing energy will find that what theylearn through taking an Applied Energy Systems module inHydrostatic Fluid Power Transmission and Flow in Pipes willserve them well. Not all applied technologists work at the

forefront of technical development and innovation. The everydaywork of society must go on. If that work involves fluids in any way,the learner will find that their learning within the scope of thepresent work will have been relevant and worthwhile. However,the relevance of the present work is significantly broader yet. Incommon with other subject areas in an engineering programme,this Applied Energy Systems topic involves engineering units(specifically the SI system) and involves important and genericengineering approaches such as experimentation, measurement,representation of practical problems in mathematical form,teamwork and communication. The present work is intentionallycompact and concise. It requires further explanation orelaboration. Learners are encouraged to make use of the Internetand library services for additional and supplementaryinformation.


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What is a Fluid?A fluid is a substance that can flow. Generally it would be a liquidor a gas. A solid is not a fluid. So, what distinguishes a fluid from anon-fluid? The following is a suitable definition:A fluid is a substance that cannot sustain a shear stresswithout undergoing relative movement.

The definition of a fluid involves the term ‘shear stress,’ whichitself requires some explanation. A stress is a force per unit area(S.I. units newton/m2). Two main categories of stresses arenormal stresses and shear stresses. If a force is normal to the areaover which it acts it is said to be a normal force and the stress istherefore a normal stress. A tensile force applied to a bar would be

an example of a normal force. If a force is in the same plane as thearea over which it acts it is called a shear force and thecorresponding stress is therefore a shear stress. Figure 1illustrates normal and shear forces and stresses. The normalstress exists because a bar is being pulled from both ends; as thebar is in tension this normal stress can be described as a tensilestress. The shear stress exists because a bar is subjected to forcesby blades of a guillotine or shears that are tending to shear the barin the plane in which the forces act, the shear plane.

Figure 1 Normal and shear forces and stresses


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Solid materials can withstand high shear forces without anyrelative movement occurring within the material. However, whena fluid is subjected to a shear stress, relative movement alwaysoccurs within the fluid, no matter how small the shear stress.Because of this characteristic, a liquid will always flow under theinfluence of gravity unless it is prevented from flowing further bya container wall below the liquid. The surface of a liquid atequilibrium within a container will be horizontal, because if itwere not horizontal, shear forces capable of causing relativemovement would exist within the liquid.


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Physical quantity SI base unit Other unitsvolume m 1 cm 10 m

1 mm

10

m

liter: 1 L 10 m milliliter:1 mL 10 m

pressure Pa (pascal) 1 Pa 1 Nm 1 kPa 10 Pa

1 MPa 10 Pa1 GPa 10 Pa1 hPa 10 Pa1 bar 10 Pa

atmosphere:1 atm 101325 Pa

stress Pa (pascal) 1 Pa 1 Nm absolute or dynamicviscosity Pa sor N ms, or kg ms 1 poise 10 Pa s

1 centipoise 10 poise 10 Pa s

kinematic viscosity ms 1 stoke 10 m s

1 centistoke 10 stoke 10 m s velocity m s acceleration m s


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Physical quantity SI base unit Other unitsdensity kg m

energy, work, heat J 1 J 1 Nm1 kJ 10 J1 MJ 10 Jpower W 1 W 1 J s

1 kW 10 Wbulk modulus of

elasticity

Pa 1 Pa 1 Nm

compressibility Pa conventionaltemperature ℃ (degree Celsius)thermodynamic orabsolute temperature K (kelvin)torque (moment) Nm

angle rad 2 rad 360°angular velocity rads revolutions per second:1 r.p.s. 2 rad s revolutions per minute:

1 r.p.m. 260 rad s angular acceleration rads

volume flow rate m s 1 L s 10 m s 1 L min 10 60 m s


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Newton’s 2nd Law and the SI Units for ForceNewton’s second law has a central role in mechanics, includingfluid mechanics. It is mentioned here in order to explain anddefine the relationship between the units for force, mass, distanceand time.Newton’s second law describes the force that must be applied to amass to cause it to accelerate. It can be summarized in words as:

Force equals mass times acceleration.

This can be expressed mathematically as (1)

where = force N = mass kg = acceleration m s As with all equations that represent physical quantities, the unitsof the left hand side of the equation must be equivalent to the

units of the right hand side of the equation. Newton’s second law,Equation 1, therefore defines the relationship between thenewton, the kilogram, the metre and the second. Of these fourunits, any one can be expressed in terms of the other three. Forexample1 N 1 kg m s

1 kg 1 N m s .Hence, wherever the unit the newton is used it can be replaced by

the combination of units kg m s

. Also, it can be deduced thatforce has the dimensions MLT. Pressure would therefore havethe dimensions MLTL, which is equivalent to ML T.


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Chapter 3.

Pressure and

Compressibility

PressurePressure is force exerted per unit area in a direction that is normalto and towards the area. It is expressed as (2)

where p

= pressure Pa or Nm −2 F = force N A = area m 2One pascal (Pa) is a very low pressure; for instance, atmosphericpressure is about 105 Pa. Absolute PressureAbsolute pressure is true pressure, which is the force per unitarea exerted on a surface. The entire force normal to the area isincluded. The lowest possible value of absolute pressure is zero(in any pressure units). Atmospheric PressureThe pressure of the atmosphere is caused by the weight of the airthat constitutes it. At sea level the atmospheric pressure equalsthe weight of air that is supported by a unit area of horizontalsurface. At higher levels less air is supported so the atmosphericpressure is correspondingly lower.

1 standard atmosphere = 1.01325 10 5 PaAtmospheric pressure can be measured by means of a traditionalmercury barometer, Figure 2, or, for instance, by an aneroidbarometer, containing a sealed cell in the form of a bellows as inFigure 3.


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ℎ (3)where

= density kg/m3 = acceleration due to gravity m s−2 ℎ = depth below free surface mIt is common to describe a pressure by quoting the correspondinghead of water or, sometimes, of other fluids, e.g. mercury (density13,546 kg m−3).Compressibility and Bulk Modulus

The compressibility of a fluid is a measure of how much its volumechanges with pressure. The volumetric change is described asvolumetric strain, i.e. the change in volume divided by the originalvolume.Suppose a fluid is compressed, as shown in the Figure 4, due to theapplication of a small additional force, F.

Figure 4 Change in volume owing to an in increase in pressureCompressibility, ∆ ∆ 1 ∆∆

(4)The term ‘bulk modulus of elasticity’ is used for the inverse ofcompressibility:


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Bulk modulus, ∆ ∆ ∆∆

(5)

Bulk modulus has the same units as pressure, i.e. Pa. The bulkmodulus of liquid water is about 2.2 GPa (1 GPa = 109 Pa), whichindicates how much pressure is required to produce volumetricstrain.All liquids have high values of bulk modulus: for hydraulic fluids itranges from about 1.4 to 1.8 GPa. Gases have very low bulkmodulus—for isothermal compression the bulk modulus isroughly equal to the pressure.

In an oil-hydraulic power system the compressibility can usuallybe ignored when sizing components. The fluid is treated as‘incompressible.’Example 1 Bulk ModulusWhat increase in pressure would be required to cause avolumetric strain of 1% in water (i.e. to reduce the volume by1%)?Solution

∆ ∆ 2.2 10

Pa 0.01 2.2 10

Pa Thus, a pressure of about 220 atmospheres would be necessary toreduce the volume of liquid water by 1%.


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Chapter 4.

Hydrostatic Work and

Power Transmission

Pascal’s LawIt is possible to talk about the pressure at a point in a fluid byconsidering the force on a very small area centred about thatpoint. Pascal’s law can be stated as follows:Pressure within a fluid at rest acts equally in

all directions and at right angles to any

surface on which it acts.

Hydraulic power transmission systems take advantage of this lawin transmitting pressure from one part of a system to another.While differences in head may exist within a hydraulic powersystem, the overall height of the system is usually such thatpressure differences between different levels within the fluid arenegligible in comparison to the operating pressures that are used.The term ‘hydrostatic machine’ can be used for a machine inwhich Pascal’s law applies to a good approximation within bodiesof connected fluid. Figure 5 illustrates how forces depend on areasin a hydrostatic system or machine.

Figure 5 Schematic diagram of a general hydrostatic machine ContinuityIn the system shown in Figure 6 an incompressible fluid iscontained between two pistons. The piston on the left moves tothe right applying a force to the fluid. The fluid applies a force tothe piston on the right causing it to move to the right and upwardsin the direction of the pipe.


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Mass and Volume Flow RatesFrom Equation 7 the volume flow rate at a given cross section of apipe is given by the product of the area and the average velocity,

. (8)If the density of the fluid is known then the mass flow rate at thecross section is given by . (9)

The fundamental SI units for mass flow rate are kg/s.Work, Flow Work and Power

Figure 6 and Figure 7 can help us to grasp the meaning of ‘flowwork.’ In Figure 6 work is done by one piston and on anotherpiston, while in Figure 7 equivalent ‘flow work’ or ‘flow power’occurs without the presence of pistons at the flow cross sectionsof interest. In a sense, the flow work for a volume of fluid thatpasses through a cross section is the energy being transferredthrough that cross section by the hydraulic fluid and the flowwork per unit time is the power being transmitted by thehydraulic fluid.

The work done on the fluid by piston 1 in Figure 6 is given by∆ Δ Δ.Similarly, the work done on piston 2 by the fluid is given by

∆ Δ Δ.From Pascal’s law and neglecting friction, the same pressure, p,acts at positions 1 and 2 and throughout the connected fluid.Hence

∆ ∆ ∆ ∆ (10)where ∆ is the volume through which each piston sweeps.Equation 10 describes the work done by the left-hand piston onthe fluid in Figure 6 and also the work done by the fluid on the


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right-hand piston. In each case the volume displacement at thepiston is Δ.If the work at the pistons occurs over a small time interval Δ then

the rate at which work is done is given by the work amountdivided by the time interval: ∆∆ ∆∆ (11)where = power W

= pressure Pa = volume flow rate m3/s = time s = volume m3 = work JExample 2 Piston Work

a.

What work is done by a piston of diameter 0.05 m when itexerts a pressure of 200 kPa and moves through a distance of0.04 m?b. If the movement occurs in 2 s, what is the average rate ofwork, or power?Solution

W AΔs 200 10 Pa 0.05 4 m 0.04 m 15.7 J answer a. Δ 15.7 J2 s 7.85 W answer b.


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Design and Operating Principles of a Hydraulic Jack

Figure 8 Schematic representation of a hydraulic jack The hydraulic jack shown in Figure 8 is in a stationary state, withthe throttle release valve closed. The oil in the space below thepiston on the right is under pressure, owing to the weight of thepiston and the load. If an operator applies downward pressure onthe lever, the plunger on the left will move downwards in thecylinder. Oil will flow towards the right through the one-way valvein the bottom pipe and will cause the load to be raised.The pressure relief valve is similar to a one-way valve, but has avery high spring force applied so that it will only open when themaximum desired system pressure is reached.If the throttle release valve is opened, the load on the right willcause the oil to flow through the restriction of the throttle releasevalve into the hydraulic oil tank. In this way the load can belowered smoothly.If the left hand plunger is in its lowest position and is caused torise by an operator lifting the lever, oil will flow into the left handcylinder from the hydraulic oil tank (or reservoir).Therefore the machine (a hand-operated hydraulic jack) allowsthe load to be raised or lowered in a controlled manner. Thepressure relief valve provides protection against over pressure.

Plunger

Lever

Air vent

Hydraulic

oil

Load

Throttle release valve

Pressure relief valve

(strong spring)

One-way valve(weak spring)


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Chapter 5.

Positive Displacement

Pumps

A positive displacement pump is a pump that discharges a fixedvolume of pressurized fluid for each rotation of the input shaft orfor each cycle in the case of a reciprocating pump. The standardhydraulic symbol for a pump is shown in Figure 9.

Figure 9

Hydraulic pump symbol

Simple Reciprocating Pump

Figure 10 Schematic representation of a hydraulic hand pump A reciprocating positive displacement pump, as shown in Figure10, discharges a fixed amount of pressurized fluid for each cycle ofthe piston. The cycle is made up of an intake stroke and adischarge stroke. The displacement of the pump per cycle is equalto the product of the stroke and the cross-sectional area of thecylinder, Equation 12.

displ cyl stroke (12)wheredispl = pump displacement per cycle


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cyl = cross-sectional area of cylinderstroke = pump strokeExample 3 Hand Pump Displacement

If the piston of a hand-operated pump has a diameter of 30 mmand a stroke of 20 mm, determine the displacement per cycle. 4 0.03

m4 0.7068 10 m Hence,

displ 0.7068 10 m 0.02 m 14.14 10 m Swash Plate Pump

The swash plate pump, Figure 11, contains a rotating piston blockthat houses a number of pistons. The piston ends press against aswash plate that causes the pistons to undergo a reciprocatingmotion. Note the following features:A swash plate, which is a flat thrust plate at an angle to the axis ofrotation of the pumpCheck valves for intake and discharge

A fixed valve plate that contains two curved slots: one todistribute the intake fluid and the other to collect the dischargefluidA piston block, pistons, a shoe plate and piston shoes, all of whichrotate with the shaftA spring that pushes the shoe plate against the swash plate andthe piston block against the valve plateA drain port so that any leakage past the pistons can be returnedto the tank of the hydraulic system


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Figure 11 Schematic diagram of a swash plate piston pump The displacement of the swash plate pump per revolution of thepump drive shaft is given by

displ cyl cyl stroke (13)wherecyl = number of cylindersRotary Positive Displacement PumpsRotary positive displacement pumps, e.g. gear pumps or vanepumps, discharge a fixed amount of pressurized fluid for eachrotation of the input shaft.Gear PumpsIn a gear pump, Figure 12, the fluid is transported from the intake

port to the discharge port. Note carefully the directions of rotationof the gears and the path followed by the fluid. The displacementof the gear pump per revolution of the drive shaft can becalculated from the dimensions and geometry of the gears and thehousing.

Discharge valve

Intake valve

Swash plate

Shoe plate

Piston shoe

Shaft

Discharge

slot

Intakeslot

Piston block (rotating)

Piston Valve plate(fixed)

Spring

Drainto tank


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Figure 12

Schematic representation of a gear pumpVane PumpA basic type of vane pump is shown schematically in Figure 13.Volumes of fluid are enclosed on the inlet side of the pump andtransported to the discharge side.As with other rotary positive displacement pumps, thedisplacement per rotation of the drive shaft can be calculatedfrom the dimensions and geometry of the housing, rotor andvanes. Note that the casing recesses in Figure 13 are necessary asthe fluid is incompressible and the enclosed volume should notchange as it is brought from the inlet to the outlet side of thepump.

Figure 13 Schematic diagram of a vane pump

VanesHousing

Rotor

Inlet Discharge

Recess

Recess


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Ideal Flow Rate of a Positive Displacement PumpThe ideal flow rate of a positive displacement pump can becalculated as the product of the displacement of the pump perrotation (or cycle) and the speed in rotations (or cycles) per unittime, ideal displ (14)wheredispl = pump displacement per revolution (or cycle) = speed (rotations or cycles per unit time)Example 4 Ideal Flow Rate of a Vane Pump

A vane pump having a displacement of 25 mL per revolution runsat a speed of 1750 r.p.m. What flow rate would it ideally produce?Express the answer in m/s and also in litres per hour.Solution

displ 25 mLrev 25 10 m

rev 1750 revs

min 1750

60

rev

s 29.17 rev

s

ideal displ 25 10 mrev 29.17 revs 729.3 10 ms

orideal 729.3 10 m

s 1000 L

m 3600s

h

2625 Lh


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Efficiency of Hydraulic Pumps

Mechanical Efficiency of a PumpThe mechanical efficiency of a pump is defined aspump, mech ideal ∆actual

(15)wherepump, mech = pump mechanical efficiencyideal = ideal volume flow rate m /s

∆ = pressure difference across pump Paactual = actual pump input power WIt is important to note that the flow rate used in the definition ofthe mechanical efficiency, Equation 15, is the ideal flow rate,based on the theoretical displacement of the pump.Volumetric Efficiency of a PumpAn ideal positive displacement pump would have a fixed

displacement and would always discharge that amount of fluid foreach rotation (or cycle, in the case of a reciprocating pump). Inpractice, there is usually some leakage backwards through thepump so that the amount of fluid discharged is less than the idealamount (corresponding to the theoretical displacement of thepump). The leakage within a pump depends on the pressuredifference: higher leakage rates occur when the pump is operatingagainst higher pressure differences.The volumetric efficiency of a pump is defined as

pump, vol actualideal (16)wherepump, vol = pump volumetric efficiency


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actual = actual volume flow rate m /sOverall Efficiency of a PumpThe overall efficiency of a pump, Equation 17, is the product of themechanical and volumetric efficiencies. Some typical values ofhydraulic pump efficiencies are shown in Table 2.pump, overall pump, mech pump, vol

ideal ∆actual actualideal actual ∆actual (17)

wherepump, overall = pump overall efficiencyTable 2 Typical hydraulic pump efficiencies Pump Type MechanicalEffic. /[%] VolumetricEffic. /[%] OverallEffic. /[%]Gear 90–95 80–90 72–86Vane 90–95 82–92 74–87Piston 90–95 90–98 81–93Example 5 Gear Pump EfficiencyA gear pump operates at 1,500 r.p.m. with a pressure difference of15 MPa and has a displacement of 12 mL per revolution. Themeasured flow rate is 0.948 m/h. The mechanical power input todrive the pump is 4.77 kW. Calculate the mechanical efficiency, thevolumetric efficiency and the overall efficiency of the pump.SolutionThe ideal flow rate of the pump is given by

ideal 12 10 m 150060 s 300 10 ms


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pump, mech ideal ∆actual

300 10 m

s 15 10 N

m

4.77 10 W 0.943 94.3%actual 0.9843600 m

s

273.3 10 ms

pump, vol actualideal 273.3300 0.911 91.1%pump, overall pump, mech pump, vol 0.943 0.911 0.859 85.9%


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Chapter 6.

Hydraulic System

Components

Check Valve and Pressure Relief ValveFigure 14 illustrates the construction of a spring loaded one-wayvalve. Figures 15 and 16 are the schematic representations ofcheck valves with and without springs respectively.

Figure 14 Simple one-way or check valve

Figure 15 Check valve symbol (with spring)

Figure 16 Check valve symbol (without spring)

Figure 17 Operating principle of a pressure relief valve Figure 17 illustrates the construction principles of a pressurerelief valve. The pressure relief valve can be adjusting by turning a

Adjustment screw

Out

In

Spring


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knob. The resulting force, exerted by the compressed spring,determines the pressure at which the valve will open. Thestandard symbol for a pressure relief valve is shown in Figure 18.

Figure 18 Pressure relief valve symbol Hydraulic Cylinders or Rams

Hydraulic cylinders or rams are linear actuators that can be usedto apply high forces and provide pushing or pulling movements. Atypical construction for a hydraulic ram is illustrated in Figure 19.

Figure 19 Typical construction of a commercial hydrauliccylinder A ram is double acting if it can apply force while extending andwhile retracting. If the ram can only apply force in one direction itis said to be single acting—some external means will be requiredto return the piston to its original position. A hydraulic ram may

have a rod that extends from one end only or may have a rod thatextends from both ends. These configurations are described as‘single rod’ and ‘double rod’ respectively. Figures 20 to 22 are theschematic representations of three possible configurations.

Figure 20 Double acting single rod cylinder

Cylinder cap

Piston

seal

Rod seal

Cylinder seal

Rod

Piston

Port

Port

Cylinder

head

Cap eye

or bearingRod eye

or bearing

Cylinder barrel


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Figure 21 Single acting cylinder

Figure 22 Double acting double rod cylinder Cylinder Force CalculationsThe main parameters of a hydraulic cylinder are the bore, thecylinder rod diameter and the stroke of the piston. These

parameters determine the volumes of hydraulic fluid required tofill the cylinder. The cylinder bore and the piston rod diameterdetermine the thrust or tension generated at a given operatingpressure.

Figure 23 Force calculation for the cap end of a cylinderFigure 23 shows the force produced by the pressure acting on thepiston at the cap end of a single-acting cylinder. It is calculated as cyl (18)where = force produced in the piston rod

= pressure cyl = cylinder cross-sectional area


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Figure 24 Force calculation for the rod end of a cylinderFigure 24 shows the force produced by the pressure acting on thepiston at the rod end of a single-acting cylinder. It is a tensile forcewithin the rod and is calculated as

cyl (19)

where rod = rod cross-sectional areaExample 6 Cylinder Forcesa. A single-rod double-acting cylinder with a bore of 160 mm isrequired to exert a force of 200 kN on the outstroke.Determine the required oil pressure in the cap end of thecylinder.

b.

If the diameter of the cylinder rod of the same double-actingcylinder is 90 mm, determine the pulling force of the rod onthe retract stroke with the same oil pressure, this time in therod end of the cylinder.Solution

cyl 4 0.16

4 m 0.02011 m

cyl 200 10 0.02011 Nm 9.945 10 Pa

9.95 MPa a.


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rod 4 0.09

4 m 0.006362 m

eff cyl rod 0.02011 m 0.00636 m 0.01375 m eff 9.945 10 Pa 0.01375 m

136,743 N 136.7 kN b.Directional Control ValvesIn order to be able to switch between extending or retracting ahydraulic cylinder, a directional control valve is required. In thevalve symbols shown in Figure 25 and Figure 26 each square boxrepresents a possible configuration of flow paths between the fourports of the valve.

Figure 25

Two position directional control valve symbol

Figure 26 Three position direction control valve Many different methods of actuating a valve are possible, some ofwhich are illustrated by the standard symbols shown in Figure 27.


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Figure 27 Symbolic representations of different valve actuationtypes Directional Spool Valve

Various designs for directional control valves are possible, but acommon arrangement makes use of a spool, which is a solid objectin the form of a series of cylinders of different diameters. Figure21 includes a schematic illustration of the construction of a threeposition, spring return, solenoid actuated directional control spoolvalve as well as the standard symbolic representation of such adevice. Letters A and B label the connections to the device beingcontrolled, such as a hydraulic ram or a hydraulic motor. P labelsthe connection from the pump and T labels the connection to the

tank.

Figure 28 Three-position directional spool valve

Solenoid

Lever

Manuallyoperated

Spring

ButtonMechanicallyoperated

BA P

T

Solenoid

Spool

BA

TP


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The Hydraulic Circuit DiagramFigure 29 is an example of a hydraulic circuit for a double actingcylinder and illustrates how a directional control valve isconnected. This circuit is made up entirely of standard symbolsfor hydraulic components such as the cylinder, the pump, thepressure relief valve, a filter and the hydraulic tank or reservoir. Itshould be noted that the tank is shown three times in the circuitdiagram, although there is only one hydraulic fluid tank. Thisavoids the clutter that would be caused if all return lines wereshown connected to just one symbol for the tank.

Figure 29

Basic hydraulic circuit for a double acting single rodcylinder Filter

Tank


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Chapter 7.

Design of Hydraulic

Systems

Control of hydraulic systems is introduced through the topic offlow control. The principles of hydraulic system design areintroduced by means of two worked examples, for a lift and apress respectively. It is important to keep in mind that in designthere are always choices and judgements to be made.Flow Control

Flow Restriction and Flow ControlThe flow of fluid in a hydraulic circuit can be reduced by means of

a flow restrictor, Figure 30. This causes a pressure drop in thedirection of flow. By means of a variable area valve, the flow ratecan be controlled. Generally, the flow rate through a restrictor isproportional to the effective flow area and to the square root ofthe pressure drop across it.

Figure 30

Symbols for a fixed area flow restrictor and a variablearea flow restrictor Power is always dissipated in a flow restrictor. This has the effectof increasing the temperature of the hydraulic fluid.Hydraulic Ram Speed ControlThe speed of a hydraulic ram can be controlled by the use of aflow control valve in conjunction with a pressure relief valve, asillustrated in Figures 31 to 33. Three possible arrangements areillustrated, named ‘meter-in,’ ‘meter out’ and ‘bleed-off’ flowcontrol.


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Figure 31 Meter-in speed control metering circuit

Figure 32 Meter-out speed control metering circuit

Figure 33 Bleed-off speed control metering circuitWith meter-in and meter-out flow control, a proportion of theflow is forced through the pressure relief valve. With meter-incontrol the force available to drive the load is greater. Meter-outcontrol creates a back pressure on the rod side of the piston whichcan help prevent lunging if the load decreases rapidly or reverses.With bleed-off control, fluid is bled-off through the restrictor at alower pressure than the pressure relief valve setting so there isless power wastage. Bleed-off control can only be used where theload opposes the motion.


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Power wastage is inevitable when metering flow control is used.An alternative would be to use a hydraulic pump whosedisplacement can be varied. Figure 34 is the standard symbol forsuch a pump.

Figure 34 Standard hydraulic symbol for a variable displacementpumpHydraulic System for a Lift

Example 7 Component Sizing for a Hydraulic Lift

A hydraulic lift is required to raise loads of up to 20 tonne througha distance of 4 m in 30 s. Select a reasonable operating pressureand use this to size a suitable cylinder and pump. Do not exceed arunning pressure of 120 bar. Determine the exact runningpressure and specify a relief valve setting. Assume an overallpump efficiency of 60% and estimate the required electric motorpower output.The following standard bores should be used:

25, 32, 40, 50, 63, 80, 100, 125, 160, 200, 250, 320 mm.SolutionFirst, to find the required cylinder bore, use ⁄ , and a trialpressure of 120 bar. 20 1,000 kg 9.81m s 196,200 N

120 10 Nm

1.96210

N120 10 N m 0.0164 m and 4 / 4 0.0164/ m 0.1445 m 145 mm.

We must choose a standard size, the nearest ones being 125 and160 mm.


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In order to keep the running pressure below 120 bar, we choosethe larger (160 mm) bore.Running pressure

run 1.962 10 N 0.1604 m 9.75810 Nm 97.6 barIt would be reasonable to set the pressure relief valve at apressure that is 10% higher than the pressure required by themaximum load. Hence, the setting would be

rv 97.6 bar 110% 107 bar .

Pump size: 0.1604 m 430 m s 0.00268 ms 2.68 Iitres 161 litremin

Electric motor size:Hydraulic power 107 10 0.00268 W

28,676 W 28.7 kWEstimated electric motor power Hydraulic output powerOverall pump efficiency 28.7 kW0.6 47.8 kW

The next standard motor power rating above this would beselected, e.g. 50 kW.


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Hydraulic PressFigure 35 is a possible circuit for a hydraulic press. With thevariable area flow restrictor fully open only a slight pressure willbe applied to the piston. The pressure can be increasedprogressively by gradually closing the restrictor.

Figure 35 Circuit for a hydraulic press Figure 36 shows how a hydraulic cylinder might be incorporatedinto a load frame of a press.

Figure 36 Load frame of a hydraulic press Example 8 Design of a Simple Hydraulic Press

A 25 tonne hydraulic press based on a single-acting spring-returncylinder is required to stroke through 300 mm in 6 seconds. If themaximum allowable pressure rating is 150 bar(a) Specify a suitable standard cylinder bore and(b) Size a suitable pump.

Vent


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The following standard bores should be considered:25, 32, 40, 50, 63, 80, 100, 125, 160, 200, 250 & 320 mm.Solution

A suitable design for the press might be as follows:(a) Specify a suitable standard bore for the cylinderThe force exerted by the press is 25 tonne weight (strictly, theweight of a mass of 25 tonne). This must be converted to newtons.1 tonne weight 1 1,000 9.81 N, where 9.81 m/s is theacceleration due to gravity. Hence

25 1,000 9.81 245,250 N.The maximum pressure of 150 bar must be converted to N/m2.Hence 150 10 Nm .The force is also equal to the product of the pressure by the areaof the piston. Hence in terms of the piston diameter or cylinderbore, d ,

4

or 4 .Hence

4 245,250 150 10 NNm 0.1443 m.

Therefore a standard bore of 0.160 m could be used. The requiredoperating pressure would therefore beop 150 [bar] × 0.14430.16 122.01 bar

The pressure relief valve would be set at perhaps 10% or 10 barhigher, whichever is the smaller. Hence the pressure relief valve


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would be set at about 132 bar. Note that 132 bar is less than thestated maximum allowable pressure rating.(b) Size a suitable pump

The volume flow rate into the cylinder is the product of thevelocity and the cross-sectional area, i.e. /4. The velocity,, is 0.36 ms 0.05 ms .

The volume flow rate is given by π 4 .

Therefore the volume rate in litres per minute is 0.05 0.164 m

s 6 0 smin 1,000 litrem 60.32 litre min.

A suitable pump would be capable of operating at up to 132 barand with a flow rate of about 61 litres per minute.


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Chapter 8.

ViscosityViscosity is a property of a fluid, which quantifies how strongly

that fluid resists relative motion. A fluid that has a high resistanceto relative motion, such as crude oil, would have a high viscosityand would be described as being viscous. A fluid that presented noresistance to relative motion would be described as inviscid, butunder normal circumstances all fluids have a finite viscosity,however small.Figure 37 illustrates a block sliding horizontally over a layer ofliquid that covers a horizontal surface. Figure 38 illustrates howthe velocity varies within the fluid layer: the fluid velocity is zeroat the stationary surface and increases linearly with distance fromthe surface to the velocity, , of the moving surface.

Figure 37 Block sliding on a film of liquid

Figure 38 Velocity profile within a fluid undergoing shear motion

It is found that the force required to maintain the velocity isproportional to the surface area of the moving block and to thevelocity , while it is inversely proportional to the distance between the stationary surface and the moving block. This can bewritten as ∝ A .

Area AVelocity v

Thickness d

Area A

Distance d

Stationary

Force F

Velocity profile

Velocity v

y

x


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This relationship of proportionality can be made into an equationby including a constant of proportionality, , as follows: A

.

(20)Equation 20 is known as Newton’s viscosity equation. Theconstant (the Greek letter mu) is a fluid property, which isknown as the absolute or dynamic viscosity.Newton’s Viscosity EquationWhereas in Figure 38 the velocity profile is a straight line, ingeneral when a fluid flows over a stationary surface the velocityprofile is likely to be represented by a curve, as shown in Figure39. It is found that Newton’s viscosity equation applies for eachthin layer of fluid, e.g. for the layer shown in Figure 39 that has athickness of Δ. Over this small thickness, the velocity of the fluidincreases by an amount that is represented by Δ.

Figure 39 General velocity profile within a fluid Newton’s viscosity equation can be written in a more general formthan Equation 20 as

Δ

Δ .

(21)At this point it is worth pausing to make sure that we understandwhat the force in Equation 21 represents. It represents theshear force, owing to viscosity, between adjacent layers of themoving fluid. The layers can be regarded as sliding, one overanother.

Moving fluid

Stationarysurface

Velocity profile

y

x

y

v


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For the velocity profile shown in Figure 39, we can see that as wemove away from the stationary surface, Δ becomes smaller forsuccessive equal steps of , the distance from the surface.Therefore Δ/Δ becomes smaller as we move away from thesurface. Hence, the shear force

F

between layers of the movingliquid decreases as we move further from the fixed surface where 0.Newton’s viscosity equation, Equation 21, gives the force that actsover a plane of area within the fluid. We can easily modify it togive the force per unit area over the same plane, which is theshear stress over the plane. We do this by dividing both sides ofthe equation by the area . Hence, the shear stress (the Greekletter tau) is given by

ΔΔ . (22)The SI Units Involved in Newton’s Viscosity EquationThe SI units for distance, velocity, area and force are relativelystraightforward, as summarized here: = distance in directionnormal to displacement m = distance between surfaces m = fluid velocity m s = force NWe can note that N stands for the SI unit of force, the newton. Wecan also note that m s−1 is another way of writing m/s, whichstands for ‘metres per second.’ As stress is defined as forcedivided by area (or force per unit area) its units are the units offorce divided by the units of area. Hence = shear stress N m Even if we do not know or remember what the units of absoluteviscosity are, we can determine the units easily from Newton’sviscosity equation, Equation 21 or Equation 22. RearrangingEquation 21,


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Δ Δ . (23)

The absolute viscosity must therefore have the same units as theright hand side of Equation 23. These units areN mmm s Nm s.Therefore the correct SI unit for absolute viscosity is Nms orPa s, which could also be written as Ns/m2. To summarize, basedon what has been covered so far, the symbol and units forabsolute viscosity are: = absolute viscosity Pa s (or Nms)Newton’s 2nd Law and the SI Units for Absolute ViscosityThe SI units for absolute viscosity can be expressed in a form thatincludes the unit for mass, the kilogram. Newton’s second law,Equation 1, Page 8, is the key to the equivalence between formsthat include mass and those that include force.

(1) (repeated)

Of the four units the newton, the kilogram, the metre and thesecond, any one can be expressed in terms of the other three. Forexample 1 N 1 kg m s.This allows a version of the SI units for absolute viscosity to beused that does not include the newton, but does include thekilogram.

1 N ms 1 kg m sms 1 kg ms.

Summarizing again, the symbol and units for absolute viscosityare: = absolute viscosity Pa s or Nms or kg ms It is important to be aware of these different ways of writing the SIunits for absolute viscosity. All three versions given above are inuse.


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Kinematic ViscosityKinematic viscosity (the Greek letter nu) is defined as

where = kinematic viscosity m2 s −1 = density ( massvolume kg/m3We can note that the units for kinematic viscosity are got bydividing the units of absolute viscosity by the units of density. Asthe density units include kg, it is convenient to use the form of theabsolute viscosity units that includes kg. Thus, the units ofkinematic viscosity are represented askg mskg m m s.More about Viscosity Units and ValuesAbsolute (or dynamic) viscosity values for real fluids, expressed inthe fundamental SI units, are usually very small, e.g. 1 10−3 Pa sfor water. Kinematic viscosity values are even smaller, e.g.1 10 m s for water. Other viscosity units are also in usesuch as1 poise = 10−1 Pa s (absolute viscosity)

1 centipoise = 10−2 poise = 10 −3 Pa s (absoluteviscosity)1 stoke = 10−4 m 2 s −1 (kinematic viscosity)1 centistoke = 10−2 stoke = 10 −6 m 2 s−1 (kinematicviscosity)


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Figure 40 Schematic diagram of a Redwood viscometer Another scale for viscosity is the Redwood scale in ‘Redwoodseconds.’ The Redwood viscosity is determined by measuring howlong it takes for a fixed amount of liquid to flow through acalibrated orifice, Figure 40. The Redwood scale is a practicalscale for comparing viscosities of different liquids, which isessentially a kinematic viscosity scale. However, the relationshipbetween the Redwood scale and a true kinematic viscosity scale isnot quite linear. Therefore, values from the Redwood scale are notused directly for engineering calculations, but a calibrationformula can be used to convert test readings from a Redwoodviscometer to kinematic viscosity.For liquids generally the viscosity decreases as the temperaturerises. For gases, the opposite is generally the case: viscosityincreases with increasing temperature. Example 9 Newton’s Viscosity EquationCalculate the force required to cause relative motion at a rate of1.7 m/s of a flat plate of area 0.36 m2 with respect to a parallel flatsurface where the plate is separated by a distance of 0.6 mm andwhere the space between the plate and the surface is filled with oilhaving an absolute viscosity of 64 centipoise. (1 centipoise = 10−2 poise = 10−3 Pa s). Include a sketch in your solution.

Thermometers

Water

Plug

Stirrer

handleLevel

pointer

OilStirrer vane

Insulation

Graduated

cylinder

Electric

heater

Calibrated

orifice


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Solution

The shear force is given by ∆∆

64 10

[N

m

s] 1.7 [m s0.0006 [m] 0.36 m

65.3 N

Area AVelocity v

Thickness d


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Chapter 9.

Pressure Losses in Pipes

and Fittings

Laminar and Turbulent FlowThe easiest way to visualise the flow of a fluid within a pipe is toimagine that the velocity of the fluid is the same over the entirecross section of flow, as illustrated in Figure 41. This is known asuniform flow.

Figure 41 Uniform flowFor uniform flow, the velocity can be calculated by dividing thevolume flow rate by the cross sectional area. (24)

In the case of a real fluid, with finite viscosity, the velocity is never

perfectly uniform: the fluid in contact with the pipe wall will havezero velocity and, generally, the fluid in the middle of the pipe willhave the highest velocity. For a fluid that has a relatively lowvelocity and a relatively high viscosity the variation of the velocityacross the cross section, which is called the velocity profile, wouldbe as shown in Figure 42. In this type of flow every particle of fluidmoves with a velocity that is parallel to the centreline of the pipe.A lamina is a thin layer. Thin cylindrical laminae of fluid can beimagined to slide over one another, the laminae near the centremoving faster than those near the edges.

Figure 42 Laminar velocity profile Where the velocity is relatively high and the fluid viscosity isrelatively low, a different type of velocity profile, as shown inFigure 43, is produced. In addition to an average velocity in the


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direction of flow, each particle of fluid moves with a chaoticvelocity that causes mixing over the cross section of the pipe.Local eddies occur and continue to change with time. Theturbulent velocity flow profile comes closer to matching a uniformflow profile than the laminar profile does.Figure 43 Turbulent velocity profileOsborne Reynolds carried out an experiment in which he was ableto visually compare laminar and turbulent flow through a glasspipe. This experiment is shown schematically in Figures 44 and45. When dye is injected from a needle into laminar flow, a thindye filament can be seen to continue over a long length within theflow, Figure 44.

Figure 44 Reynolds’ experiment with laminar flow

Figure 45 Reynolds’ experiment with turbulent flow When dye is injected into a high flow rate of fluid, Figure 45, thedye filament is maintained for a certain length from the pipe inlet(indicating laminar flow), but, further on, turbulent flow developsand causes mixing of the dye over the full bore of the pipe.In laminar flow, Figure 44, viscous forces dominate, whereas inturbulent flow inertia forces (i.e. acceleration or decelerationforces) dominate. In laminar flow there is no cross mixingbetween the layers. In turbulent flow a thorough cross-mixing of

Laminar flow at very low velocity

Velocity

profile in pipeDye injection

Dye injection

Turbulent flow

at high velocity

after entry length

Velocity

profile in pipe


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the fluid occurs and individual small volumes of fluid have chaoticmotion superimposed on their average forward velocity.It is worth noting at this point that, irrespective of the type of flow,

the average fluid velocity is given by Equation 24.Reynolds NumberWhether the flow in a pipe is laminar or turbulent depends on thedensity of the fluid, the velocity of flow, the diameter of the pipeand the absolute viscosity of the fluid. These parameters can becombined in a dimensionless number, called the Reynoldsnumber. The value of this number can be used to predict the typeof flow that will exist. The Reynolds number is defined as

Re

(25)whereRe = Reynolds number = density kg/m = average fluid velocity m s

= diameter m = absolute viscosity Pa s (or Nm sor kg ms) = kinematic viscosity m s Whether the flow is laminar or turbulent depends on the Reynoldsnumber, as follows:

Laminar flow exists for Re below 2,000.Turbulent flow exists for Re greater than 4,000.Between these values laminar or turbulent flow is possible.

Example 10 Laminar or Turbulent Flow?It takes 30 seconds to fill a kettle with two litres of water from atap. The tap is supplied by a pipe that has an internal diameter of


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12.7 mm. Determine whether the flow is laminar or turbulent inthe supply pipe while the kettle is being filled. Take the density ofwater as 1,000 kg/m and take its absolute viscosity as1.002 10 kg/m s.SolutionCalculate the volume flow rate as

2 10m30 s 66.67 10 m/s.

Calculate the cross sectional area as 4 0.0127

4 m 126.7 10 m.

Calculate the mean velocity as 66.67 10m ⁄ 126.7 10m 0.5262 ms .

Calculate the Reynolds number asRe

1000 kgm0.5262ms 0.0127 m

1.002 10

kgm s 6,669 ≅ 6.7 10

As Re > 4,000, flow is turbulent.Head Loss Due to FrictionDifferent laws of fluid resistance apply for laminar and turbulentflows, although one common equation (the Darcy-Weisbachequation, which is sometimes just called the Darcy equation) canbe used for fluid flow in pipes—it incorporates a ‘friction factor’that depends on the nature of the flow.


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The Darcy-Weisbach equation has the formℎf 4

2

(26)where the friction factor f is obtained as detailed below and whereℎf = head loss due to friction m = friction factor = length of pipe m = diameter m

= mean fluid velocity m s−1 = acceleration due to gravity m s−2 Note: The friction factor used here is known as the Fanningfriction factor. In some textbooks and references the Darcy-Weisbach equation is written without the integer 4 in thenumerator, but the friction factor used is four times the frictionfactor used here. That friction factor is known as the Darcy frictionfactor.

As might be expected, the internal roughness of a pipe, known asits surface roughness, can have an influence on the pressure dropthat occurs in the pipe. In fact, it is the ratio of the surfaceroughness to the diameter that is of importance. This ratio, knownas the relative roughness, influences the pressure drop forturbulent flow. However, for laminar flow the relative roughnessdoes not influence the pressure drop.

Figure 46 Diameter and surface roughness of a pipe

d k


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For turbulent flow the friction factor f is commonly obtained froma friction factor diagram, which is often referred to as a Moodydiagram or chart, Figure 47. The friction factor depends on theReynolds number and on the relative roughness of the pipe k /d where = Pipe surface roughness m = Pipe internal diameter mThe friction factor can also be found by using appropriateempirical formulae, e.g. within a spreadsheet. These formulae aretoo cumbersome for hand calculations; hence the friction factordiagram is useful.

For fully turbulent flow the head loss per unit length of pipe, ℎf ⁄ ,is proportional to the mean velocity squared and depends on therelative roughness of the pipe. For flow that is not fully turbulentthere is not a direct proportionality to velocity squared (roughlyspeaking, the head loss is proportional to velocity raised to apower that is close to and a little lower than 2).For laminar flow the friction factor is given by

16Re .

(27)Note 1: Equation 27 is written on the friction factor diagram, sothere is no need to remember it.Note 2: In references where the friction factor is four times thevalue used here Equation 27 is written with 64 in the numeratorinstead of 16.It can be shown that for laminar flow the head loss per unit lengthof pipe, ℎf ⁄ , is proportional to the mean velocity (or the meanflow rate) and independent of surface roughness. The frictionfactor for the laminar flow region can also be read from a frictionfactor diagram.


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Figure 47 Friction factor diagram (Moody diagram).

1 0 3

1 . 4

2

3

4

5 6

1 0 4

1 . 4

2

3

4

5 6

1 0 5

1 . 4 2

3

4

5 6

1 0 6

1 . 4

2

3

4

5 6

1 0 7

1 . 4

2

3

4

5 6

1 0 8

1 . 4

2

3

4

5 6

1 0 9

0 . 0 0 1 3

0 . 0 0 1 4

0 . 0 0 1 6

0 . 0 0 1 8

0 . 0 0 2

0 . 0 0 2 2

0 . 0 0 2 4

0 . 0 0 2 6

0 . 0 0 2 8

0 . 0 0 3

0 . 0 0 3 5

0 . 0 0 4

0 . 0 0 4 5

0 . 0 0 5

0 . 0 0 6

0 . 0 0 7

0 . 0 0 8

0 . 0 0 9

0 . 0 1

0 . 0 1 2

0 . 0 1 4

0 . 0 1 6

0 . 0 1 8

0 . 0 2

1 0 3

1 . 4

2

3

4

5 6

1 0 4

1 . 4

2

3

4

5 6

1 0 5

1 . 4 2

3

4

5 6

1 0 6

1 . 4

2

3

4

5 6

1 0 7

1 . 4

2

3

4

5 6

1 0 8

1 . 4

2

3

4

5 6

1 0 9

0 . 0 0 1 3

0 . 0 0 1 4

0 . 0 0 1 6

0 . 0 0 1 8

0 . 0 0 2

0 . 0 0 2 2

0 . 0 0 2 4

0 . 0 0 2 6

0 . 0 0 2 8

0 . 0 0 3

0 . 0 0 3 5

0 . 0 0 4

0 . 0 0 4 5

0 . 0 0 5

0 . 0 0 6

0 . 0 0 7

0 . 0 0 8

0 . 0 0 9

0 . 0 1

0 . 0 1 2

0 . 0 1 4

0 . 0 1 6

0 . 0 1 8

0 . 0 2

V e r s i o n F M 0 0 1 2 0 1 1 . 1 0 . 0 2

R e

1 6 0 0

D

f f

1 . 0 1

a n d

R e y

n o l d s N u m b e r R e

V D

V D

F r i c t i o n F a c t o r f F h f 4 L

D

V 2

2 g p 4 L

D

V 2

2 w V

2

2

R e y n o l d s N u m b e r

F r i c t i o n F a c t o r

R e l a t i v e R o u g h n e s s 5 . 1 0 7

0 . 0 0 0 0 0 1

0 . 0 0 0 0 0 2

0 . 0 0 0 0 0 3

0 . 0 0 0 0 0 5

0 . 0 0 0 0 1

0 . 0 0 0 0 2

0 . 0 0 0 0 3

0 . 0 0 0 0 5

0 . 0 0 0 1

0 . 0 0 0 2

0 . 0 0 0 3

0 . 0 0 0 4

0 . 0 0 0 6

0 . 0 0 0 8

0 . 0 0 1

0 . 0 0 1 5

0 . 0 0 2

0 . 0 0 3

0 . 0 0 4

0 . 0 0 5

0 . 0 0 6

0 . 0 0 8

0 . 0 1

0 . 0 1 2

0 . 0 1 4

0 . 0 1 6

0 . 0 1 8

0 . 0 2

0 . 0 2 5

0 . 0 3

0 . 0 3 5

0 . 0 4

0 . 0 4 5

0 . 0 5

0 . 0 5 5

0 . 0 6

J i m M c G o v e r n , D u b l i n I n s t i t u t e o f T e c h n o l o g y

h t t p : / / w w w . f u n - e n g i n e e r i n g . n e t / p i p e f r i c t i o n f a c t o r . h t m l

F a n n i n g F r i c t i o n F a c t o r f o r P i p e F l o w

C o m p l e t e T u r b u l e n c e Z o n e

T r a n s i t i o n Z o n e

S m o o t h

P i p

e

f F 1 6 R e

L a m i n a r

Z o n e

R e l a t i v e R o u g h n e s s D

C r i t i c a l

Z o

n e

I n d i c a t i v e R o u g h n e s s V a l u e s

m m

S m o o t h h o n e d s t e e l

0 . 0

0 0 6 5

D r a w n t u b i n g : g l a s s , b r a s s , c o p p e r , l e a d , p l a s t i c

0 . 0

0 1 5

A s p h a l t e d c a s t i r o n

0 . 1

2

G a l v a n i z e d s t e e l

0 . 1

5

W o o d s t a v e

0 . 1

8 0 . 9 1

C a s t i r o n

0 . 2

6

C o n c r e t e

0 . 3

3

H e a v y b r u s h c o a t : a s p h a l t s , e n a m e l s , t a r s

0 . 4

5 0 . 6

G e n e r a l t u b e r c u l a t i o n 1 3 m m

0 . 6

1 . 9

R i v e t e d s t e e l

0 . 9

9

S e v e r e t u b e r c u l a t i o n a n d i n c r u s t a t i o n

2 . 5

6 . 5


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Flow through Valves and FittingsThe presence of a valve in a pipe, even when it is fully open,almost always causes a head loss owing to disruption of the flowand the formation of a vena contracta (reduced flow area). Thesame is true for many other types of fittings, e.g. a fitting that joinspipes of different diameters. In these cases a head loss coefficientis used, Equation 28.

ℎ 2 . (28)In the above expression the velocity is the mean velocity in thefull bore pipe upstream.K is a dimensionless constant for a particular configuration andwill depend on the nature of the valve or fitting. Hence, if the headloss at a valve or fitting is measured for a given mean velocity offlow the head loss coefficient can be determined as

2ℎ .Flow Around Bends and ElbowsWhenever the direction of flow is abruptly changed at a bend orelbow, a loss of head occurs due to the formation of a venacontracta on passing the bend or elbow and the turbulenceassociated with the subsequent enlargement. Surface roughnesswithin the bend also has an influence on the head loss. The headloss can be characterized by a head loss coefficient, as in Equation28.Equivalent Pipe LengthCommercially, the head loss caused by flow through fittings issometimes presented as an ‘equivalent length of straight pipe’ to

cause the same head loss. In this case the head loss for a pipe runthat includes fittings (elbows, bends, reducers etc.) can becalculated using the Darcy-Weisbach equation (Equation 26), butthe pipe length used is the actual length of the pipe plus theequivalent lengths of any fittings used. Head losses associatedwith valves, fittings and bends are often described as minor pipelosses. Major losses are those associated with fluid friction overthe full length of a pipe.


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Pressure Loss and Pumping PowerThe Darcy-Weisbach equation, Equation 26, allows us to calculatethe head loss due to friction in a pipe. Head loss has the units oflength, the metre. If we require the pressure loss, or pressuredrop, over the length of pipe we can use Equation 3, ℎ, tofind it: loss ℎ f (29)whereloss pressure loss due to friction Pa density kg/m acceleration due to gravity m/s ℎf head loss due to friction mCombining Equations 26 and 29

loss 4

2 or

loss 4

2 (30)Equation 30 is an alternative form of the Darcy-Weisbachequation, which gives the pressure loss rather than the head lossdue to fluid friction.In some cases flow through a pipe is caused directly by adifference in liquid levels in two tanks at either end of the pipe.The flow might also be caused directly by a pump, which wouldhave to provide a pressure difference to match the pressure lossin the pipe. The pressure difference across the pump is oftenrepresented by Δ, Figure 48. If a pump is required to exactlymeet a pressure drop due to friction in a pipe then Δ loss.


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Figure 48 Pressure difference across a pump The ideal pumping power equals the product of the pressuredifference across the pump and the volume flow rate, Equation 31. Δ (31)

where

ideal pumping power WΔ pressure difference Pa volume flow rate m/sSummary of Key Formulae for Pipe FlowThe key formulae for pipe flow are summarized in Table 3.

Table 3 Summary of key formulae for pipe flow

Mean velocity of flow in a pipe: Reynolds Number:Re Re < 2,000 Laminar Flow

Re > 4,000 Turbulent FlowHead loss due to friction in a pipe Darcy-Weisbach equation:

ℎf 4

2For laminar flow can be calculated from 16Re .For turbulent flow, f depends on Re and on the relativeroughness k /d .


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Head losses at bends or fittingsℎ 2 .Pumping power

Δ.Example 11 Head Losses due to FrictionA pipe of absolute roughness 0.0030 mm and bore 50 mm conveyswater at a rate of 210 litres per minute. If the viscosity of thewater is 0.0013 kg/m s determine(a) the friction factor(b) the head loss per metre run of pipe

(c) the maximum allowable flow rate if the flow were to belaminar(d) the theoretical pumping power per metre length for theturbulent flow in the pipe.SolutionThe volume flow rate is given by

210

60 1,000 m /s = 0.0035 m/s.

The cross-sectional area is 0.05 4 m 1.963 10 m .

The mean velocity of flow in the pipe is 0.0035 m

/s]1.96310 m ] 1.783 m/s .

The Reynolds number isRe 1,0001.783 0.050.0013 kg m m s mkg m s

68,577 ≅ 6.9 10 As Re > 4,000, the flow is turbulent.


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The relative roughness is 0.003 mm50 mm 0.00006

For Re = 68,600 and k /d = 0.00006 the friction factor is foundfrom the friction factor diagram to be 0.0049 (see diagram onpage 59). (a)The head loss due to friction in a pipe (Darcy-Weisbach equation)is:

ℎf 4

2

Hence ℎf 4

2 4 0.00490.050 1.783

2 9.81 m s m m s

0.0635 m/mThat is, the head loss is 63.5 mm per metre of pipe length. (b)The maximum allowable flow rate if the flow were to be laminarwould require Re to equal 2,000. From the definition of theReynolds number Re ) the maximum laminar flow velocityis

Re 2,000 0.00131,0000.05 kg m s kg m3m

0.0520 m/sQ vA 0.0520 m/s 1.963 10 m

0.1021 10 m /s 0.102160 L/min 6.13 L/min (c)The pumping power for the turbulent flow case is given by

∆ ℎf


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10

3

1 . 4

2

3

4

5

6

10

4

1 . 4

2

3

4

5

6

10

5

1 . 4

2

3

4

5

6

10

6

1 . 4

2

3

4

5

6

10

7

1 . 4

2

3

4

5

6

10

8

1 . 4

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